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Chapter 12: Q60 E (page 708)

The rate constant for the decomposition of acetaldehyde, \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\), to methane, \({\bf{C}}{{\bf{H}}_{\bf{4}}}\), and carbon monoxide, CO, in the gas phase is 1.1 脳 10鈭�2 L/mol/s at 703 K and 4.95 L/mol/s at 865 K. Determine the activation energy for this decomposition.

Short Answer

Expert verified

The energy of activation for the reaction is 96.769 kJ/mol

Step by step solution

01

Step 1: Using Arrhenius Equation

From the Arrhenius Equation, the rate of reaction at two different temperatures is given as

\(\log \frac{{{k_2}}}{{{k_1}}} = \frac{{{E_a}}}{{2.303*R}}\left( {\frac{1}{{{T_1}}} - \frac{1}{{{T_2}}}} \right)\)

where k1and k2are the rate constants at \({{\bf{T}}_{\bf{1}}}\) and \({{\bf{T}}_{\bf{2}}}\) where \({{\bf{T}}_{\bf{1}}}{\bf{ < }}{{\bf{T}}_{\bf{2}}}\). \({{\bf{E}}_{\bf{a}}}\)is the activation energy (in J) and R is the gas constant

02

Calculation

Replacing the values,

\(\log \frac{{4.95}}{{1.1*{{10}^{ - 2}}}} = \frac{{{E_a}}}{{2.303*8.314}}\left( {\frac{1}{{703}} - \frac{1}{{865}}} \right)\)

\({E_a} = 190691.9J/mol\)

\({E_a} = 190.69kJ/mol\)

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Most popular questions from this chapter

Based on the diagrams in Exercise 12.83, which of the reactions has the fastest rate? Which has the slowest rate?

Nitrosyl chloride, NOCl, decomposes to NO and \({\bf{C}}{{\bf{l}}_{\bf{2}}}\).

\({\bf{2NOCl(g)}} \to {\bf{2NO(g) + C}}{{\bf{l}}_{\bf{2}}}{\bf{(g)}}\)

Determine the rate law, the rate constant, and the overall order for this reaction from the following data:

Acetaldehyde decomposes when heated to yield methane and carbon monoxide according to the equation: \({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)(g) 鉄禱({\bf{C}}{{\bf{H}}_{\bf{4}}}\)(g) +\({\bf{CO}}\)(g)

Determine the rate law and the rate constant for the reaction from the following experimental data:

Trial

(\({\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}\)) (mol/L)

\(\frac{{ - \Delta \left( {{\bf{C}}{{\bf{H}}_{\bf{3}}}{\bf{CHO}}} \right)}}{{\Delta t}}\)(mol )(Ls鈭�1)

1.

1.75 脳 10鈭�3

2.06 脳 10鈭�11

2.

3.50 脳 10鈭�3

8.24 脳 10鈭�11

3.

7.00 脳 10鈭�3

3.30 脳 10鈭�10

What is the difference between average rate, initial rate, and instantaneous rate?

Experiments were conducted to study the rate of the reaction represented by this equation.(2)\({\rm{2NO(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{(g) }} \to {\rm{ }}{{\rm{N}}_{\rm{2}}}{\rm{(g) + 2}}{{\rm{H}}_{\rm{2}}}{\rm{O(g)}}\)Initial concentrations and rates of reaction are given here.

Experiment Initial Concentration

\(\left( {{\bf{NO}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)

Initial Concentration, \(\left( {{{\bf{H}}_{\bf{2}}}} \right){\rm{ }}\left( {{\bf{mol}}/{\bf{L}}} \right)\)Initial Rate of Formation of \({{\bf{N}}_{\bf{2}}}{\rm{ }}\left( {{\bf{mol}}/{\bf{L}}{\rm{ }}{\bf{min}}} \right)\)
1\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{1}}.{\bf{8}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
2\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{3}}.{\bf{6}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
3\({\bf{0}}.{\bf{00}}1{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{0}}.{\bf{30}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)
4\({\bf{0}}.{\bf{00}}2{\bf{0}}\)\({\bf{0}}.{\bf{0060}}\)\({\bf{1}}.{\bf{2}} \times {\bf{1}}{{\bf{0}}^{ - {\bf{4}}}}\)

Consider the following questions:(a) Determine the order for each of the reactants, \({\bf{NO}}\) and \({{\bf{H}}_{\bf{2}}}\), from the data given and show your reasoning.(b) Write the overall rate law for the reaction.(c) Calculate the value of the rate constant, k, for the reaction. Include units.(d) For experiment 2, calculate the concentration of \({\bf{NO}}\)remaining when exactly one-half of the original amount of \({{\bf{H}}_{\bf{2}}}\) had been consumed.(e) The following sequence of elementary steps is a proposed mechanism for the reaction.Step 1:Step 2:Step 3:Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.
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